How do you write the complex conjugate of the complex number #(3-i )/( 4-i )#?

1 Answer
Jun 23, 2016

Answer:

#bar((3-i)-:(4-i))=13/17+1/17i#

Explanation:

Given a complex number #a+bi#, that number's complex conjugate, denoted #bar(a+bi)#, is given by

#bar(a+bi) = a-bi#

To find the complex conjugate of #(3-i)/(4-i)#, we can either simplify first and then take the conjugate, or use the fact that if #z_1,z_2inCC#, then #bar(z_1-:z_2)=bar(z_1)-:bar(z_2)# and simplify afterwards.

Using the latter approach:

#bar((3-i)-:(4-i)) = bar(3-i)-:bar(4-i)#

#=(3+i)/(4+i)#

#=((3+i)(4-i))/((4+i)(4-i))#

#=(13+i)/17#

#=13/17+1/17i#

Note that we were able to perform the division by multiplying the numerator and denominator by the complex conjugate of the denominator. This method takes advantage of the property that the product of a complex number and its conjugate is a real number.